<think>
To solve the equation
$$
\sin(x)^{11} + \cos(x)^{11} = 1,
$$
we need to find values of $x$ such that the given equality holds. Recall that the sine and cosine functions have well-known properties such that:
$$
\sin^2(x) + \cos^2(x) = 1.
This implies that both $\sin(x)$$ and $\cos(x)$ must be in the interval $[-1, 1]$.

However, because the question uses only the 11th power here (which is an odd integer), any negative values for $\sin(x)$ or $\cos(x)$ would keep the sign of the original functions (i.e., $\sin(x)^{11}$ will keep the sign of $\sin(x)$ if $\sin(x) < 0$, similarly for $\cos(x)^{11}$ if $\cos(x) < 0$). Furthermore, if $|\sin(x)| < 1$, then it follows that $|\sin(x)|^{11} < \sin^2(x)$ because $|\sin(x)|^{11} < \sin^2(x)$ for $|\sin(x)| < 1$. Similarly, if $|\cos(x)| < 1$, then $|\cos(x)|^{11} < \cos^2(x)$ for $|\cos(x)| < 1$.

To make a more rigorous reasoning, let us define two new variables $s = |\sin(x)|$ and $c = |\cos(x)|$, such that $0 \leq s, c \leq 1$ where $s^2 + c^2 = 1$. Our goal is to solve for $s^{11} + c^{11} \leq s^2 + c^2 = 1$, where equality only holds if $s^{11} = s^2$ and $c^{11} = c^2$ simultaneously (since $s^{11} + c^{11} \leq s^2 + c^{2}$ whenever $s, c \leq 1$).

The only cases where $s^{11} = s^2$ are when $s = 0, 1,$ or (since $0 \leq s \leq 1$), the same applies for $c^{11} = c^2$ when $c = 0, 1$. However, since $s^2 + c^2 = 1$, it follows that if $s^2 = 1$ (i.e., $s = 1$, given that $s \geq 0$), $c$ must be $0$, and vice versa.

Another important part is to make a direct comparison for a general $s$ and $c$ such that $s, c \in [0,1]$.

For $0 \leq a < 1$, of any $k > l \geq 1$, it holds that $a^k < a^l$. Thus, for $s, c \in [0,1]$, and $s^2 + c^2 = 1$:
$$
\sin(x)^{11} + \cos(x)^{11} \leq |\sin(x)|^{11} + |\cos(x)|^{11}.
$$
However $|\sin(x)|^{11} \leq \sin^2(x)$ if $|\sin(x)| \leq 1$, which is always true unless $|\sin(x)| = 1$ where $|\sin(x)|^{11} = \sin^2(x) = 1$, or $|\sin(x)| = 0$ where $0^{11} = 0^2 = 0$.

Similarly, for $|\cos(x)|^{11} \leq \cos^2(x)$, which holds true when $0 < |\cos(x)| < 1$, and equality holds when $|\cos(x)| = 1$ or $0$.

Let us analyze what happens for $s,c \in [0,1]$ where $0 < s,c < 1$:
$$
|\sin(x)|^{11} \leq \sin^2(x) \text{ (since $|\sin x| \leq 1$, thus $|\sin(x)|^{11} \leq |\sin(x)|^2 = \sin^2(x)$)}.
$$
$$
|\cos(x)|^{11} \leq \cos^2(x) \text{ (similarly $|\cos(x)|^{11} \leq \cos^2(x))$}.
$$
This implies:
$$
|\sin(x)^{11} + \cos(x)^{11}| \leq |\sin(x)|^{11}+|\cos(x)|^{11}.
$$
However, by the triangle inequality:
$$
|\sin(x)^{11} + \cos(x)^{11}| \leq |\sin(x)|^{11} + | \cos( x )|^{11} \leq \sin^2(x) + \cos^2(x) = 1.
$$
The only way that $\sin(x)^{11} + \cos(x)^{11}$ could be equal to $1$ is if $|\sin(x)|^{11} + |\cos(x)|^{11} = 1$, which only happens when $|\sin(x)|^{11} = |\sin(x)|^2$ and $|\cos(x)|^{11} = |\cos(x)|^2$. Taking into account that $|\sin(x)|^{11} + |\cos(x)|^{11} \leq |\sin(x)|^2 + |\cos(x)|^2$, the only cases where $t^{11} = t^2$ (for $0 \leq t \leq 1$) are when $t \in \{0, 1\}$.

Thus, solutions where $|\sin(x)|^{11} + |\cos(x)|^{11} = 1$ rely on one (or both) of $|\sin(x)|$ or $|\cos(x)|$ being $0$ or $1$. Recall that $|\sin(x)|$ and $|\cos(x)|$ should satisfy $s^2 + c^2 = 1$, so $s$ and $c$ cannot both be equal to $1$. Thus, the only valid combinations are:
1. $|\sin(x)| = 1$ and $|\cos(x)| = 0$, which gives the case where $s=1$, $c=0$ such that $s^{11} + c^{11} = 1$.
2. $|\sin(x)| = 0$ and $|\cos(x)| = 1$, which gives the case where $s=0$, $c=1$ such that $s^{11} + c^{11} = 1$.

To solve $|\sin(x)| = 1$ and $|\cos(x)| = 0$, we need:
$$
\sin(x) = \pm 1 \text{ and } \cos(x) = 0.
$$
This happens when $x = \frac{\pi}{2} + k\pi$ for any integer $k$. However, we need to verify what happens separately when $\sin(x) = \pm 1$:
1. If $\sin(x) = 1$, $\cos(x) = 0$ such that:
$$
\sin(x)^{11} + \cos(x)^{11} = (1)^{11} + (0)^{11} = 1 + 0 = 1,
$$
so $x = \frac{\pi}{2} + 2k\pi$ works (since $1^{11} + 0^{11} = 1$).
2. If $\sin(x) = -1$, $\cos(x) = 0$ such that:
$$
\sin(x)^{11} + \cos(x)^{11} = (-1)^{11} + (0)^{11} = -1 + 0 = -1,
$$
so $x = \frac{3\pi}{2} + 2k\pi$ (which corresponds to $x = \frac{\pi}{2} + (2k+1)\pi)$ does not work because it yields $-1$ instead of $1$.

Conversely, for $|\sin(x)| = 0$ and $|\cos(x)| = 1$, that is:
$$
\sin(x) = 0 \text{ and } \cos(x) = \pm 1.
$$
This happens for $x = k\pi$ for any integer $k$. However, analyzing separately for the cases $x = 2k\pi$ (which makes $\cos(x) = 1$) and $x = (2k + 1)\pi$ (which makes $\cos(x) = -1$):
1. If $\sin(x) = 0$, $\cos(x) = 1$ such that:
$$
\sin(x)^{11} + \cos(x)^{11} = (0)^{11} + (1)^{11} = 0 + 1 = 1,
$$
so $x = 2k\pi$ works.
2. If $\sin(x) = 0$, $\cos(x) = -1$ such that:
$$
\sin(x)^{11} + \cos(x)^{11} = (0)^{11} + (-1)^{11} = 0 - 1 = -1,
$$
so $x = (2k + 1)\pi$ does not work because that yields $0 - 1 = -1$.

Thus, what we need is to verify what values make a $\sin(x) = 1,0$ where $\cos(x) = 0,1$, but only such that the sum makes $1$.

In order to make a more organized solution, let us consider all the cases explicitly.

Case 1: $\sin(x) = 1$ and thus $\cos(x) = 0$.
$$
\sin(x) = 1 \text{ implies }x = \frac{\pi}{2} + 2k\pi.
$$
These values make the equation valid:
$$
\begin{cases}
\sin(x) = 1 : \sin(x) = \sin\left(\frac{\pi}{2} + 2k\pi\right) = 1 \\
\cos(x) = 0 : \cos(x) = \cos\left(\frac{\pi}{2} + 2k\pi\right) = 0.
\end{cases}
$$
Thus:
$$
\sin(x)^{11} + \cos(x)^{11} = 1^{11} + 0^{11} = 1 + 0 = 1.
$$
So $x = \frac{\pi}{2} + 2k\pi$ (where $k$ is any integer) is a set of solutions.

Case 2: $\sin(x) = 0$ and thus $\cos(x) = 1$.
$$
\cos(x) = 1 \text{ implies } x = 2k\pi.
$$
These values make:
$$
\begin{cases}
\sin(x) = 0 : \sin(x) = \sin(2k\pi) = 0 \\
\cos(x) = 1 : \cos(x) = \cos(2k\pi) = 1.
\end{cases}
$$
Thus:
$$
\sin(x)^{11} + \cos(x)^{11} = 0^{11} + 1^{11} = 0 + 1 = 1.
$$
So $x = 2k\pi$ is another set of solutions where $k$ is any integer.

However, any other scenario where $|\sin(x)|,|\cos(x)| < 1$ makes the inequality strict:
$$
\sin(x)^{11} + \cos(x)^{11} < \sin^2(x) + \cos^2(x) = 1.
$$
Hence the only valid solutions appear to be $x = 2k\pi$ or $x= \frac{\pi}{2} + 2k\pi$.

However, this reasoning might need us to make a substitution $\sin(x) = u$, $\cos(x) = v$, such that $u^{11} + v^{11} = 1$ is verified along with the Pythagorean identity $u^2 + v^2 = 1$. Given $u^2 + v^2 = 1$, if we make $u = \sin \theta$ and $v = \cos \theta$, the equation becomes:
$$
\sin^{11}\theta + \cos^{11}\theta = 1,
$$
which is what we need to solve. However, we need to verify once again if $u = \sin \theta$, $v = \cos \theta$ might have any other values such that $u^{11} + v^{11} = 1$ while $u^2 + v^2 = 1$.

However, it looks like a well-known inequality for $0 < s,c < 1$ is that:
$$
s^{11} < s^2,
$$$$
c^{11} < c^2.
$$

Hence to make a more concrete argument for $0 < s < 1$ where $s = |\sin(x)|$:
$$
\sin^2 (x) > |\sin(x)|^{11} \text{ unless } |\sin(x)| \in \{0, 1\}.
$$
And for $0 < c < 1$ where $c = |\cos(x)|$:
$$
\cos^2(x) > |\cos(x)|^{11} \text{ unless } |\cos(x)| \in \{0, 1\}.
$$
Thus:
$$
\sin(x)^{11} + \cos(x)^{11} \leq |\sin(x)|^{11} + |\cos(x)|^{11} \leq \sin^2(x) + \cos^2(x) = 1,
$$
with equality if and only if one of $s$ and $c$ is $1$ while the other is $0$.

Let us solve the elementary trigonometric equations for when one of $\sin(x)$ and $\cos(x)$ is $0$ or $1$ while the other makes its absolute value $0$ or $1$.

1. $\sin(x) = 1 \Rightarrow \sin^2(x) = 1 \Rightarrow \cos^2(x) = 1 - \sin^2(x) = 0 \Rightarrow \cos(x) = 0$ (only valid case here where the sum of the $11$th powers is $1^{11} + 0^{11} = 1$).

So $x = \frac{\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$.

2. $\cos(x) = 1 \Rightarrow \cos^2(x) = 1 \Rightarrow \sin^2(x) = 1 - \cos^2(x) = 0 \Rightarrow \sin(x) = 0$ (again, only valid case here where the sum of the $11$th powers is $0^{11} + 1^{11} = 1$).

So $x = 2k\pi$, $k \in \mathbb{Z}$.

3. $\sin(x) = -1 \Rightarrow \sin^2(x) = 1 \Rightarrow \cos^2(x) = 0 \Rightarrow \cos(x) = 0$, however:
$$
\sin(x) = -1 \text{ and } \cos(x) = 0 \text{ makes } \sin(x)^{11} + \cos(x)^{11} = (-1)^{11} + 0^{11} = -1,
$$
thus this case does not satisfy the given equation.

4. $\cos(x) = -1 \Rightarrow \cos^2(x) = 1 \Rightarrow \sin^2(x) = 0 \Rightarrow \sin(x) = 0$, however:
$$
\sin(x) = 0 \text{ and } \cos(x) = -1 \text{ makes } \sin(x)^{11} + \cos(x)^{11} = 0^{11} + (-1)^{11} = -1,
$$
thus this case does not satisfy the given equation either.

So the only valid combinations where $\sin(x)^{11} + \cos(x)^{11} = 1$ are when $\sin(x)$ and $\cos(x)$ are either $0$ or $1$, such that one is $1$ while the other is $0$, which happens for:
$$
\sin(x) = 1 \text{ and } \cos(x) = 0,
$$
which corresponds to $x = \frac{\pi}{2} + 2k\pi$ for some $k \in \mathbb{Z}$, or:
$$
\sin(x) = 0 \text{ and } \cos(x) = 1,
$$
which corresponds to $x = 2k\pi$ for some $k \in \mathbb{Z}$.

Thus, the two families of solutions are:
$$
x = 2k\pi, \text{ where } k \in \mathbb{Z};
$$
$$
x = \frac{\pi}{2} + 2k\pi, \text{ where } k \in \mathbb{Z}.
$$
This means $x = 2k\pi$ or $x = \frac{\pi}{2} + 2k\pi$ for $k \in \mathbb{Z}$ can be combined into a single expression if one wishes, by using the set notation $x \in \{2k\pi : k \in \mathbb{Z}\} \cup \left\{\frac{\pi}{2} + 2k\pi : k \in \mathbb{Z}\right\}$.

Another possible way of writing these solutions separately:
$$
x = 0 + 2k\pi \text{ where $k$ is an integer, or } x = \frac{\pi}{2} + 2k\pi \text{ where $k$ is an integer.}
$$

To verify the reasoning explicitly:
$$
x = 2k\pi \text{ gives } \sin(x) = 0, \cos(x) = 1, \text{ so, } \sin(x)^{11} + \cos(x)^{11} = (0)^{11} + (1)^{11} = 0 + 1 = 1.
$$
$$
x = \frac{\pi}{2} + 2k\pi \text{ gives } \sin(x) = 1, \cos(x) = 0, \text{ so, } \sin(x)^{11} + \cos(x)^{11} = (1)^{11} + (0)^{11} = 1 + 0 = 1.
$$
These appear to be the only valid cases.

However, a bit of a double-check can be done graphically. Plot $y=\sin^{11}(x) + \cos^{11}(x)$ and $y=1$ simultaneously on a graph. We should only see intersections where one of $\sin(x)$ or $\cos(x)$ is close to $1$ (while the other is close to $0$), which reinforces our previous reasoning that the question indeed appears to be solved correctly.

Thus, the only valid solutions are:
$$
\boxed{x = 2k\pi} \text{ where } k \in \mathbb{Z} \text{ or }  \boxed{x = \frac{\pi}{2} + 2k\pi}\text{ where } k \in \mathbb{Z}.
$$

This means that the only valid solutions are:
$$
x \in \left\{{2k\pi : k \in \mathbb{Z}}\right\} \cup \left\{\frac{\pi}{2} + 2k\pi : k\in\mathbb{Z} \right\}.
$$
Alternatively, this is the same as:
$$
x \in \left\{2k\pi, \frac{\pi}{2} + 2k\pi \text{ where } k\in \mathbb{Z}\right\}.
$$
### Verification in another compact form
It makes sense to verify this reasoning explicitly within a given period such as $0 \leq x < 2\pi$:
1. At $x = 0$ (equivalent to $2k\pi$ for $k=0$ in general form):
$$
\sin(0) = 0, \cos(0) = 1,
$$
$$
\sin(0)^{11} + \cos(0)^{11} = 0^{11} + 1^{11} = 0 + 1 = 1.
$$
$x=0$ is a valid solution.
2. At $x = \frac{\pi}{2}$ (equivalent to $\frac{\pi}{2} + 2k\pi$ for $k=0$ in general form):
$$
\sin\left(\frac{\pi}{2}\right) = 1, \cos\left(\frac{\pi}{2}\right) = 0,
$$
$$
\sin\left(\frac{\pi}{2}\right)^{11} + \cos\left(\frac{\pi}{2}\right)^{11} = 1^{11} + 0^{11} = 1 + 0 =1.
$$
$x = \frac{\pi}{2}$ is a valid solution.
3. At $x = \pi$ (equivalent to $\pi = (2k+1)\pi$ for $k=0$ in general form):
$$
\sin(\pi) = 0, \cos(\pi) = -1.
$$
$$
\sin(\pi)^{11} + \cos(\pi)^{11] = 0^{11} + (-1)^{11} = 0 - 1=-1.
$$
$x = \pi$ is not a valid solution.
4. At $x = \frac{3\pi}{2}$:
$$
\sin\left(\frac{3\pi}{2}\right) = -1, \cos\left(\frac{3\pi}{2}\right) = 0.
$$
$$
\sin\left (\frac{3\pi}{2}\right)^{11} + \cos\left(\frac{3\pi}{2}\right)^{11} = (-1)^{11} + 0^{11} = -1 + 0 = -1.
$$
$x = \frac{3\pi}{2}$ is not a valid solution.

This reasoning makes us confident that the only values within one round ($0 \leq x < 2\pi$) that make $\sin(x)^{11} + \cos(x)^{11} = 1$ are indeed $x = 0$ (which is equivalent to $x=2k\pi$ for any $k \in \mathbb{Z}$) or $x = \frac{\pi}{2}$ (which is equivalent to $x = \frac{\pi}{2} + 2k\pi$ for any $k \in \mathbb{Z}$).

Considering any other values of $x$, such that $\sin(x) \neq 1$ or $0$, or $\cos(x) \neq 1$ or $0$, one has the inequality $|\sin(x)| < 1$ or $|\cos(x)| < 1$, which makes the left-hand side $\sin(x)^{11} + \cos(x)^{11} < 1$.

Therefore, the only solutions are thus:
$$
\boxed{x = 2k\pi} \text{ for } k \in \mathbb{Z}, \text{ or } \boxed{x = \frac{\pi}{2} + 2k\pi} \text{ for }  k \in \mathbb{Z}.
$$
Verified.
</think>

\boxed{x = 2k\pi} \text{ where } k \in \mathbb{Z} \text{ or } x = \frac{\pi}{2} + 2k\pi \text{ where } k \in \mathbb{Z}